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14y^2+16y-30=0
a = 14; b = 16; c = -30;
Δ = b2-4ac
Δ = 162-4·14·(-30)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1936}=44$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-44}{2*14}=\frac{-60}{28} =-2+1/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+44}{2*14}=\frac{28}{28} =1 $
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